Newton's Law exam Questions

Main ideas / concepts covered

The video solves a multi-part mechanics question using Newton’s Laws, with emphasis on Newton’s second law.

It highlights:

• Newton’s first law (equilibrium): An object is in equilibrium only if it has zero acceleration (i.e., it is at rest or moves with constant velocity).

• Newton’s second law (dynamics): When the net force is non-zero, the object accelerates in the direction of the net force: [ F_{\text{net}} = ma ]

• Free-body diagrams (FBDs): How to draw them and which forces to include.

• Friction and kinetic friction: Friction opposes motion. Kinetic friction depends on the coefficient of friction and the normal force.

• Tension in a light string over a frictionless pulley: Tension is the same throughout the string (same magnitude along the rope segments).

Methodology / step-by-step instruction (as used in the questions)

1) Define Newton’s second law and set up equations

• Use:
  • Net force in a chosen direction equals mass × acceleration.
• Choose positive directions consistent with expected motion (often right for the 8 kg block).

2) Draw labeled free-body diagrams (FBDs)

For the 8 kg block on a horizontal rough surface, include:

• Weight: (W = mg) downward
• Normal reaction: (N) upward
• Friction: (F) opposing motion (direction chosen based on the block’s tendency to move)
• Tension: (T) from the string, including its horizontal/angled effect

For the 2 kg hanging block, include:

• Weight (W = mg) downward
• Tension (T) upward along the string
• No normal force (since it’s not on a surface in the hanging setup)

Note from the speaker: For the CAPS curriculum, they advise not using component labels on FBDs (even though components may still appear in some memos).

3) Use Newton’s second law for the 2 kg mass to find tension

• Write vertical forces on the 2 kg mass (with the downward direction as positive, as described): [ W - T = ma ]

• Substitute (W = 2 \cdot 9.8) and (a = 1.32).

• Solve for tension.

The stated result is:

• [ T = 16.96\ \text{N} ] The speaker later uses 1696-style numbers, and the same tension magnitude is used consistently for the friction calculation as described.

4) Use Newton’s second law for the 8 kg block to find kinetic friction

• Apply horizontal dynamics: [ T\cos(15^\circ) - f_k = ma ]

• Rearrange: [ f_k = T\cos(15^\circ) - ma ]

• Substitute (m = 8), (a = 1.32), and the previously found (T).

The stated result is:

• [ f_k = 5.82\ \text{N} ] acting left (direction chosen to oppose motion; the positive calculation confirms the assumed direction).

5) Answer reasoning-based qualitative questions

• Why not in equilibrium? Because acceleration is not zero, so forces are not balanced.

• Why kinetic friction is not constant from B to C? As the block moves, the angle between the string and the horizontal changes (it increases), which changes the effective horizontal tension component and thus affects friction behavior.

• Whether friction changes if the surface material changes Yes, because kinetic friction depends on: [ f_k = \mu_k N ] Changing the material changes the coefficient (\mu_k), so the kinetic friction force changes.

Main results explicitly stated

• Reason it’s not equilibrium: acceleration (\neq 0).
• Tension in the string: stated as 1696 N (used in subsequent calculations).
• Kinetic friction force on the 8 kg block: 5.82 N, acting left.
• Why friction is not constant from B to C: the string angle to the horizontal changes (increases).
• How friction changes with surface material: because (\mu_k) changes.

Speakers / sources

• Speaker: an unnamed instructor/teacher (the person narrating the solutions).